Freefall from Infinity

Assuming a system with relatively small but non-negligable mass falling from infinity to a maximally massive black hole;
Would there be a maximum possible coordinate velocity below c ????
Is it possible to do an approximate calculation for this?
A ballpark figure??
If there is a limit, what is the physics behind it?

A related question : Is there a minimal dilation factor for the clock of a Schwarzschild observer at infinity, independant of the mass of the gravitating body???

Assuming a system with relatively small but non-negligable mass falling from infinity to a maximally massive black hole;
Would there be a maximum possible coordinate velocity below c ????
Is it possible to do an approximate calculation for this?
A ballpark figure??
If there is a limit, what is the physics behind it?

A related question : Is there a minimal dilation factor for the clock of a Schwarzschild observer at infinity, independant of the mass of the gravitating body???

Thanks

The local velocity of a free falling object as measured by a stationary observer at r for an object that is initially at rest and released from infinity is:

[tex]v= c \sqrt{\frac{Rs}{r}} [/tex]

For an observer very close to the event horizon so that r is nearly equal to Rs, the local velocity approaches c. The catch is that we can not have a (*stationary*) observer exactly at (*or below*) the event horizon (r=Rs) so the velocity is never exactly c but we can take the observer as close as we like to the event horizon and the maximum velocity to as close as we like to c as long as the difference we are satisfied with is not exactly zero.

So you can have the maximum velocity as 0.99c or 0.9999999999999999c or as many 9's after the decimal point as you like, but you can't have v=c.

The time dilation factor of a stationary clock in the Schwarzschild metric is [itex]d\tau = dt\sqrt{(1-Rs/r)}[/itex]. If you move away from the gravitational source so that r tends towards infinite, then the time dilation factor tends to unity, but since in practice you probably never arrive "exactly" at infinity, the time dilation factor of the "distant clock" is probably never exactly unity, but you take as close to unity as you like.

The local velocity of a free falling object as measured by a stationary observer at r for an object that is initially at rest and released from infinity is:

[tex]v= c \sqrt{\frac{Rs}{r}} [/tex]

For an observer very close to the event horizon so that r is nearly equal to Rs, the local velocity approaches c. The catch is that we can not have an observer exactly at the event horizon (r=Rs) so the velocity is never exactly c but we can take the observer as close as we like to the event horizon and the maximum velocity to as close as we like to c as long as the difference we are satisfied with is not exactly zero.

I fully agree with yuiop here.

Since you asked about the coordinate velocity falling from infinity I will add this here.

The coordinate velocity does not have a maximum but this is a good opportunity to address a misunderstanding that many people have. A gravitational field does not necessarily accelerate an object towards the center of gravity. In fact above a certain velocity gravitation decelerates it, by the way, just like in the case of light.

See the attached graph for the coordinate velocity of 3 different scenarios (initial velocity=0, 1/sqrt(2) and 1/sqrt(3) ), you can see what I am saying.

Now you mention that the object in question has mass, for simplicity I assume you mean that in that case the object does not have to travel at c (by the way the formula I wrote above also works for c). If not then the problem becomes a lot more complex. Is that what you want to discuss?

What is the velocity profile of a free falling object inside the event horizon?

There are no stationary observers inside the horizon, so you can't define something like an absolute velocity.
yuiop's formula is also valid inside the horizon, if you redefine it to mean [itex]dr/d\tau[/itex], radial coordinate distance / proper time. That's not a velocity, though.